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Planck integrals from Fourier Parseval

Statement

Assume the Axiom of Countable Choice. The following improper integrals converge and have the exact values $\int_{0}^{\infty} \frac{x^{3}}{e^{x} - 1} d x = \frac{π^{4}}{15}, \int_{0}^{\infty} x^{2} [- \log (1 - e^{- x})] d x = \frac{π^{4}}{45} .$

Facts & Assumptions

Given: Countable Choice and the coordinate characters on the normalized one-dimensional torus $T = R / Z$ .

[A1]

All integrals in the Statement and Proof are nonnegative Lebesgue integrals on $(0, \infty)$ with respect to one-dimensional Lebesgue measure, as defined in [F4]. The integrands and partial sums below are continuous, hence Borel measurable.

[F1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): every countable sequence of nonempty sets has a choice function; this is the exact set-theoretic assumption in the Fourier supplier.

[F2]

The Fourier basis and Parseval's identity on the finite torus: under Countable Choice, the characters $e_{n} (y) = e^{2 π i n y}$ form an orthonormal basis of $L^{2} (T)$ and Parseval gives $∥ F ∥_{2}^{2} = \sum_{n \in Z} ∣ \hat{F} (n) ∣^{2}$ .

[F3]

Monotone convergence for the integral, Statement: if nonnegative measurable functions increase pointwise to $f$ on a measure space, their integrals increase to $\int f$ .

[F4]

The nonnegative Lebesgue integral, Definition: the integral of a nonnegative measurable function is the supremum of the integrals of its nonnegative simple minorants.

Proof

1.1givenalgebra

On $[- \frac{1}{2}, \frac{1}{2}]$ define the continuous periodic representative $F (y) = (2 π y)^{2}$ . Its normalized Fourier coefficients are $c_{n} = (2 π)^{- 1} \int_{- π}^{π} x^{2} e^{- i n x} d x$ after the change $x = 2 π y$ . Direct integration gives $c_{0} = π^{2} / 3$ ; for $n \neq 0$ , two integrations by parts give $c_{n} = 2 (- 1)^{n} / n^{2}$ because the first boundary term cancels at $x = \pm π$ and $\int_{- π}^{π} e^{- i n x} d x = 0$ .

2.1F1F2step 1.1algebra

The normalized squared norm is $(2 π)^{- 1} \int_{- π}^{π} x^{4} d x = π^{4} / 5$ . Under the stated Countable Choice assumption [F1], Parseval [F2] therefore gives $π^{4} / 5 = ∣ c_{0} ∣^{2} + \sum_{n \neq 0} ∣ c_{n} ∣^{2} = π^{4} / 9 + 8 \sum_{n = 1}^{\infty} n^{- 4}$ , whence $\sum_{n = 1}^{\infty} n^{- 4} = π^{4} / 90$ .

3.1A1F3F4step 2.1algebra

For $m \geq 1$ and $p = 0, 1, 2, 3$ , put $J_{p} (m) = \int_{0}^{\infty} x^{p} e^{- m x} d x$ . Direct integration gives $J_{0} (m) = 1 / m$ ; integration by parts on $[0, R]$ and then $R \to \infty$ gives $J_{p} (m) = p J_{p - 1} (m) / m$ , since $R^{p} e^{- m R} \to 0$ (for example, $e^{m R} \geq (m R)^{p + 1} / (p + 1)!$ ). Hence $J_{3} (m) = 6 / m^{4}$ and $J_{2} (m) = 2 / m^{3}$ . The finite geometric identity and its limit give $(e^{x} - 1)^{- 1} = \sum_{m \geq 1} e^{- m x}$ for $x > 0$ . The partial sums $x^{3} \sum_{m \leq N} e^{- m x}$ are nonnegative measurable and increase pointwise; the monotone-convergence theorem [F3] for the integral defined in [F4] gives the first integral as $\sum_{m \geq 1} J_{3} (m) = 6 \sum_{m \geq 1} m^{- 4} = π^{4} / 15$ .

4.1A1F3F4step 2.1step 3.1algebra∎

For $0 < z < 1$ , the geometric series is uniformly convergent on $[0, z]$ , so integrating its partial sums gives $- \log (1 - z) = \sum_{m \geq 1} z^{m} / m$ . At $z = e^{- x}$ the nonnegative partial sums $x^{2} \sum_{m \leq N} e^{- m x} / m$ increase pointwise to the second integrand. Applying [F3] to the integral defined in [F4] and using step 3.1 gives its integral as $\sum_{m \geq 1} J_{2} (m) / m = 2 \sum_{m \geq 1} m^{- 4} = π^{4} / 45$ . The integral ratio is therefore $1 / 3$ .

Planck integrals from Fourier Parseval

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources